(Note: This post builds on work from this previous MSE question.)
Let $\sigma(N)$ denote the sum of the divisors of the natural number $N$. If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. Denote the abundancy index $I$ of $N$ as $I(N)=\sigma(N)/N$.
Euler proved that every odd perfect number $N$ has to have the form $N=q^k n^2$ where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
In this paper, Dris proved that we have the bounds $$L(q) = \frac{3q^2 - 4q + 2}{q(q-1)} = 3 - \frac{q-2}{q(q-1)} < I_{+} \leq 3 - \frac{q-1}{q(q+1)} = \frac{3q^2 + 2q + 1}{q(q+1)} = U(q)$$ for the sum of the abundancy indices $I_{+}=I(q^k)+I(n^2)$. In particular, if $I_{+} \leq 3 - \epsilon$, then we will have $L(q) < 3 - \epsilon$, so that (using WolframAlpha) we get $$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon}$$ if $0 < \epsilon < 3 - 2\sqrt{2}$.
Here is the problem:
PROBLEM STATEMENT
Does the converse hold? That is, if $N=q^k n^2$ is an odd perfect number, does $$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon},$$ with $0 < \epsilon < 3 - 2\sqrt{2}$, imply that $I(q^k) + I(n^2) \leq 3 - \epsilon$?
MY ATTEMPT
Assume that $$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon},$$ with $0 < \epsilon < 3 - 2\sqrt{2}$, and suppose to the contrary that $$I(q^k) + I(n^2) > 3 - \epsilon.$$
This means that $U(q) > 3 - \epsilon$, so that (by using WolframAlpha again) we obtain (for $0 < \epsilon < 3 - 2\sqrt{2}$) either $$q > \frac{\bigg(\sqrt{\frac{1}{{\epsilon}^2} - \frac{6}{\epsilon} + 1} - 1\bigg)\epsilon + 1}{2\epsilon},$$ or $$0 < q < \frac{1-\epsilon}{2\epsilon} - \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}.$$
Here is my question:
Do the resulting inequalities $$q > \frac{\bigg(\sqrt{\frac{1}{{\epsilon}^2} - \frac{6}{\epsilon} + 1} - 1\bigg)\epsilon + 1}{2\epsilon},$$ or $$0 < q < \frac{1-\epsilon}{2\epsilon} - \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}},$$ for $0 < \epsilon < 3 - 2\sqrt{2}$, contradict the assumption $$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon}?$$